Solving simple differential equation (t+4)dx=4(1+x^{2})dt

killjoy1990xb9

killjoy1990xb9

Answered question

2022-01-20

Solving simple differential equation
(t+4)dx=4(1+x2)dt

Answer & Explanation

Bob Huerta

Bob Huerta

Beginner2022-01-20Added 41 answers

As you rearranged it in your post, we have
dx1+x2=4dtt+4
As you discovered, the LHS is arctan. To find the value of the RHS, we may say t+4=u so du=dt. Thus the RHS can be rewritten as
4duu=4log|u|+C2
Because we want this integral to be in respect to t, we have u=t+4, so
arctanx+C1=4log|t+4|+C2
To confirm that the integral we solved is indeed correct, you can simply differentiate 4log|t+4|+C2
Thus x=tan(4log|t+4|+C2C1)
and t=exp(14(arctanx+C1C2))4
Dawn Neal

Dawn Neal

Beginner2022-01-21Added 35 answers

Here are some proceeding steps:
- Seperating the variables you get
dx1+x2=4dtt+4
- Integrating both sides you have:
tan1(x)+C1=4log|t+4|+C2.

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Differential Equations

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?